package dp;

/**
 * @author pengfei.hpf
 * @date 2020/3/17
 * @verdion 1.0.0
 * 72. 编辑距离
 * 给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。
 *
 * 你可以对一个单词进行如下三种操作：
 *
 * 插入一个字符
 * 删除一个字符
 * 替换一个字符
 * 示例 1:
 *
 * 输入: word1 = "horse", word2 = "ros"
 * 输出: 3
 * 解释:
 * horse -> rorse (将 'h' 替换为 'r')
 * rorse -> rose (删除 'r')
 * rose -> ros (删除 'e')
 * 示例 2:
 *
 * 输入: word1 = "intention", word2 = "execution"
 * 输出: 5
 * 解释:
 * intention -> inention (删除 't')
 * inention -> enention (将 'i' 替换为 'e')
 * enention -> exention (将 'n' 替换为 'x')
 * exention -> exection (将 'n' 替换为 'c')
 * exection -> execution (插入 'u')
 */
public class MinDistance {
    public int minDistance(String word1, String word2) {
        if(word1 == null){
            return word2 == null? 0: word2.length();
        }
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        dp[0][0]=0;
        //TODO: 别忘了是长度作为dp, 所以最后应该要到length
        for(int i = 1; i <= word1.length(); i++){
            dp[i][0] = i;
        }
        for(int i = 1; i <= word2.length(); i ++){
            dp[0][i] = i;
        }
        for(int i = 1; i <= word1.length(); i++){
            for(int j = 1; j <= word2.length(); j ++){
                char c1 = word1.charAt(i - 1);
                char c2 = word2.charAt(j - 1);
                if(c1 == c2){
                    dp[i][j] = dp[i -1][j - 1];
                } else {
                    dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i-1][j-1], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }

    public static void main(String[] args){
        System.out.println(new MinDistance().minDistance("i", ""));
    }
}
